# Time: O(logn)
# Space: O(1)
#
# Reverse digits of an integer.
# 
# Example1: x = 123, return 321
# Example2: x = -123, return -321
# 
# click to show spoilers.
# 
# Have you thought about this?
# Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!
# 
# If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.
# 
# Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, 
# then the reverse of 1000000003 overflows. How should you handle such cases?
# 
# Throw an exception? Good, but what if throwing an exception is not an option? 
# You would then have to re-design the function (ie, add an extra parameter).
#

class Solution:
    # @return an integer
    def reverse(self, x):
        ans = 0
        if x >= 0:
            while x:
                ans = ans * 10 + x % 10
                x /= 10
            return ans
        else:
            return -self.reverse(-x)
        
if __name__ == "__main__":
    print Solution().reverse(123)
    print Solution().reverse(-321)